# URGENT: Differential calculus

• Dec 14th 2007, 02:29 PM
Rydbirk
URGENT: Differential calculus
Okay, my english is not perfect, but I hope you can understand what I'm talking about.

I'm having this subject about movement and air resistance. We're watching an object dropping in an area with air resistance. The resulting force on the object is (using Newton's 2. law):
F_res = -F_grav + F_air <=> ma = -mg + k*v^2

Let's introduce s(t) (don't know what it's called in english), and thereby s'(t) = v(t) and s''(t) = a(t).

After some paraphrasing, we get a(t) = k/m * v^2 - g <=> s''(t) = C * (s'(t))^2 - g with C = k/m

To find v(t) and s(t) we integrate, so
v(t) = 1/3 * C * t * (s(t))^3 - g * t + v_0 and
s(t) = 1/12 * C * t^2 * ((int.)s(t)dt)^4 - g * t^2 + v_0*t + s_0

Is this correct? How do I find s(t)??? My calculator can't find it.. Maybe I'm typing something wrong or I've done something wrong.. It's really urgent, please help!
• Dec 14th 2007, 04:27 PM
Rydbirk
D'oh!
I made a mistake! You can't integrate like that..

$s'(t)=\int{(s'(t))^2dt}*C-g*t+v_{0}$
That makes it easier. Well, this is my first time using seperation ^^ so I might return..
• Dec 14th 2007, 05:34 PM
topsquark
Quote:

Originally Posted by Rydbirk
Okay, my english is not perfect, but I hope you can understand what I'm talking about.

I'm having this subject about movement and air resistance. We're watching an object dropping in an area with air resistance. The resulting force on the object is (using Newton's 2. law):
F_res = -F_grav + F_air <=> ma = -mg + k*v^2

Let's introduce s(t) (don't know what it's called in english), and thereby s'(t) = v(t) and s''(t) = a(t).

After some paraphrasing, we get a(t) = k/m * v^2 - g <=> s''(t) = C * (s'(t))^2 - g with C = k/m

To find v(t) and s(t) we integrate, so
v(t) = 1/3 * C * t * (s(t))^3 - g * t + v_0 and
s(t) = 1/12 * C * t^2 * ((int.)s(t)dt)^4 - g * t^2 + v_0*t + s_0

Is this correct? How do I find s(t)??? My calculator can't find it.. Maybe I'm typing something wrong or I've done something wrong.. It's really urgent, please help!

I think your English is pretty good. (Handshake)

s(t) is called the "displacement." It roughly corresponds to "position."

No, your work is not correct. You have a differential equation
$s''(t) = C * (s'(t))^2 - g$

But
$\int (s'(t))^2~dt \neq \frac{1}{3}(s'(t))^3$

You need to first write this as a velocity equation:
$v'(t) = C(v(t))^2 - g$

At the moment I can't remember how to solve this problem. Perhaps one of the other Helpers will post on this.

-Dan
• Dec 14th 2007, 05:42 PM
Rydbirk
Yes
Yes, I've concluded that as well - but thanks ;) And thank you for the information.

But now I'm stuck again. I've spent 2-3 hours working with this and I've recieved a lot of help, but I haven't been tought how to deal with separation in differential calculations - so I beg you, please, somebody, show me how to solve the following step-by-step and explain to me, how you do. Please, it's very important!

$s''(t) = C*(s'(t))^2-g$
• Dec 14th 2007, 05:59 PM
topsquark
Quote:

Originally Posted by topsquark
$v'(t) = C(v(t))^2 - g$

I'm an idiot. I just remembered how to do this.

$\frac{dv}{dt} = Cv^2 - g$

$\frac{dv}{Cv^2 - g} = dt$

$\int \frac{dv}{Cv^2 - g} = \int~dt$

$\frac{1}{2\sqrt{Cg}}~ln \left ( \frac{v\sqrt{C} - \sqrt{g}}{v\sqrt{C} + \sqrt{g}} \right ) = t + A$
where A is the integration constant, set by the initial conditions.

As you can see this is not a nice equation to solve for v, but it can be done:
$v(t) = -\sqrt{\frac{g}{C}}~\frac{exp[2t \sqrt{Cg} + 2A \sqrt{Cg}] + 1}{exp[2t\sqrt{Cg} + 2A\sqrt{Cg}] - 1}$

I'll leave it to you to integrate this to get s(t). (This is even less pretty, but still doable.)

-Dan
• Dec 14th 2007, 06:13 PM
Rydbirk
Thanks!!
Topsquark, thank you thank you thank you!!

So A must be my $v_{0}$.

There just 2 steps I don't get, though..
How do you isolate $\frac{1}{2\sqrt{Cg}}$ ? And how do you go from (4) to (5) ?

If noone has the time to help me out, I'll return in the morning - in Denmark it's 3 AM ;)
• Dec 14th 2007, 06:25 PM
topsquark
Quote:

Originally Posted by Rydbirk
Topsquark, thank you thank you thank you!!

So A must be my $v_{0}$.

There just 2 steps I don't get, though..
How do you isolate $\frac{1}{2\sqrt{Cg}}$ ? And how do you go from (4) to (5) ?

If noone has the time to help me out, I'll return in the morning - in Denmark it's 3 AM ;)

$\int \frac{dx}{x^2 - a^2} = \int \left ( \frac{A}{x + a} + \frac{B}{x - a} \right )~dx$
using partial fractions. Find A and B and you can integrate term by term.

The constant out in the front can be found by changing the integration variable from v to $u = \sqrt{C} \cdot v$. The rest of the constant comes from the partial fraction coefficients A and B.

-Dan
• Dec 17th 2007, 05:37 AM
Rydbirk
See this thread for further help:
http://www.mathhelpforum.com/math-he...-calculus.html

:o
• Dec 18th 2007, 02:37 PM
Rydbirk
Hmmm...
My calculator gives me a different solution:

$
\frac{1}{2\sqrt{Cg}}~ln \left ( \frac{\sqrt{g} + v\sqrt{C}}{\sqrt{g} - v\sqrt{C}} \right ) = t + A
$

But why?